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Course Code: MEG 205
Course Title: ENGINEERING MECHNICS I
Unite: 3
Course Content
ENGINEERING MECHANICS I (STATICS) Fundamentals of mechanics; Forces in space, equivalent system, equilibrium of rigid bodies, distributed forces, center of gravity, internal actions, analysis of simple structures and machine parts. Learning outcomes: On successful completion of this course students will be able to: Recall trigonometric laws and apply to the addition and decomposition of vectors quantities. Identify the moment of a force and calculate its value about a specified axis. Define the moment of a couple. Describe the concept of dry friction and analyze the equilibrium of rigid bodies subjected to this force. Construct "Free Body Diagrams" of real world problems and apply Newton's Laws of motion and vector operations to evaluate equilibrium of particles and bodies. Apply the principles of equilibrium of particles and bodies to analyze the forces in planar truss members. Discuss the concepts of center of gravity'' andcentroids'' and compute their location for bodies of arbitrary shape. Apply the concepts used for determining canter of gravity and centroids to find the resultant of a generally distributed loading Use methods learnt for equilibrium of bodies and the resultant of a generally distributed loading to compute the internal forces in beams. Generalize the procedure to construct bending moments and shear force diagrams (internal forces) and utilize this information in engineering design.
Texts: (1) Engineering Mechanics_Static (Meriam and Kraige) (2) Engineering Mechanics 1 (Gross et al.)
Note
OLABISI ONABANJOUNIVERSITY, AGO - IWOYE
COLLEGE OF ENGINEERING & ENVIRONMENTAL STUDIES IBOGUN CAMPUS FACULTY OF ENGINEERING
DEPARTMENT OF MECHANICAL
INTRODUCTION
The tasks of mechanics include the description and determination of the motion of bodies, as well as the investigation of the forces associated with the motion. Example wheel of a vehicle rolling, flight of an airplane Mechanical quantities such as velocity, mass, force, momentum or energy describing the mechanical properties of a system are connected with mechanics. Bodies may be considered rigid if their deformation does not really affect their mechanical properties.
Concept for mechanics: Space: geometric region occupied by bodies whose positions are described by linear and angular measurements Time : measure of the succession of events Mass: quantity of matter in a body or measure of the inertia of a body
Force Physical quantity that can be brought into equilibrium with gravity. Characterized by three properties: magnitude , direction , and point of application. (vector quantity)
(source: Gross et al.)
A Particle : body of negligible dimensions (dimension of approximately zero). Rigid body : when the change in distance between any two of its points is negligible (e.g. tension in a rope).
Scales an vectors Scalar - magnitude is associated (e.g. time, volume, density, speed, energy, mass) Vector - direction as well as magnitude (displacement, velocity, acceleration, force, moment, and momentum)
- iv) Determine the deference D = V1 - V
iv) Vector difference D
Force System
Complete description of forces: must include i) Magnitude, ii) direction and iii) point of application. Action of forces are either external (Applied forces or reactive forces) or internal. Principle of transmissibility : force may be applied at any point on its given line of action without altering the resultant effects of the force external to the rigid body on which it acts. We need only to specify magnitude, direction, and line of action of the force.
Example The forces F1, F2, and F3, all of which act on point A of the bracket, are specified in three different ways. Determine the x and y scalar components of each of the three forces.
OR
The resultant can also be written as:
Example The 500-N force F is applied to the vertical pole as shown. (1) Write F in terms of the unit vectors i and j and identify both its vector and scalar components. (2) Determine the scalar components of the force vector F along the x- and y-axes. (3) Determine the scalar components of F along the x- and y-axes.
Scalar components are Fx = 250 N and Fy = 433 N. The vector components are Fx = 250i N and Fy = 433j N
F = 500 i ’ N
Completing the parallelogram, thus we have
The required scalar component are: Fx = 1kN ; Fy’ = -866 N
Assignment: A homogeneous beam (length 4a, weight W) is suspended at C by a rope. The beam touches the smooth vertical walls at A and B. Find the force in the rope and the contact forces at A and B.
The unit vector nOB along OB is
Equilibrium
When a body is in equilibrium, the resultant of all forces must be zero. R = ∑F = 0 M ∑M = 0
Once we decide which body or combination of bodies to analyze, we then treat this body or combination as a single body isolated from all surrounding bodies. The isolation is accomplished by means of the free-body diagram (FBD).
Key concepts for FBD
- Decide which system to isolate. The system must include the unknown.
- Isolate the chosen system by drawing a diagram which represents its complete external boundary.
- Identify all forces which act on the isolated system as applied by the removed contacting and attracting bodies, and represent them in their proper positions on the diagram of the isolated system.
- Show the coordinate axes directly on the diagram and the dimensions.
Modeling action of forces
Example Determine the magnitudes of the forces C and T, which, along with the other three forces shown, act on the bridge-truss joint.
To avoid a simultaneous solution, we may use axes x’ – y’ with the first summation in the y’-direction to eliminate reference to T
Example
The uniform 7-m steel shaft has a mass of 200 kg and is supported by a ball-and-socket joint at A in the horizontal floor. The ball end B rests against the smooth vertical walls as shown. Compute the forces exerted by the walls and the floor on the ends of the shaft.
FBD:
W = mg = 200(9.81) = 1962 N
The vertical position of B is found from:
Use A as a moment center to eliminate reference to the forces at A.
To calculate the forces at A:
OR
Scalar solution:
Question 3
Structures
Examples of trusses
Statically determinate structures: free-body diagram is essential for the analysis of statically determinate structures.
In trusses, additional members or supports that are not necessary but are just for maintaining the equilibrium configuration are called redundant. For simple trusses analysis: we assume two-force members. The two forces are applied at the ends of the member and are equal , opposite, and collinear for equilibrium.
Free body diagram
Calculate the force in each member of the load cantilever truss by the method of joints.
FBD:
For Joint A:
For Joint B: ( Why point B next?)
Point C:
Exercise: Determine the force in each member of the loaded truss.
Method of sections
Space truss is the Three-dimensional of the plane truss
